Question

Resolva os problemas:

Derive implicitamente : $y^{3} +y=x$.

Use a regra de L'Hospital e calcule o limite: $\lim_{nx\to +\infty} \frac{e^{x} }{x} ;$

Answer 1

a)

$y^3+y=x\\\\ \frac{d}{dx} (y^3+y) = \frac{d}{dx} (x)\\\\ \frac{d}{dx}(y^3) + \frac{d}{dx}(y) = \frac{d}{dx}(x)\\\\ 3y^2 \frac{dy}{dx} + \frac{dy}{dx}= \frac{dx}{dx}\\\\ 3y^2 \frac{dy}{dx}+\frac{dy}{dx} =1\\\\ \frac{dy}{dx}(3y^2+1)=1\\\\ \frac{dy}{dx}= \frac{1}{3y^2+1}$


b)

$\lim_{x\to +\infty} \frac{e^{x} }{x} = \lim_{x\to +\infty} \frac{(e^{x})' }{(x)'} = \lim_{x\to +\infty} \frac{e^{x} }{1} = \lim_{x\to +\infty} e^{x} = e^{\infty} = \infty$